Another rather simple Differential Equation

[JoshHolloway]

$$y \prime = (cos^{2}x)(cos^{2}(2y))$$

$$\frac{dy}{dx} = \frac{1}{2} (cox(2x) + 1) \frac{1}{4}(cos(4y) + 1)$$

$$\frac{1}{cos(4y) + 1}dy \ = \ \frac{1}{8} (cos(2x) + 1)dx$$

$$\int \frac{1}{ \sqrt{cos(4y)}^{2} + 1}dy \ = \ \frac{1}{8} \int (cos(2x) + 1)dx$$

$$tan^{-1}(cos4y) \ = \ \frac{1}{8} \frac{1}{2} sin(2x) + \frac{1}{8} x + C$$

$$tan[tan^{-1}(cos4y) \ = \ \frac{1}{16} sin(2x) + \frac{1}{8} x + C]$$

$$cos^{-1}[cos(4y) \ = \ tan( \frac{sin2x}{16} + \frac{x}{8} + C)]$$

$$y \ = \ \frac{1}{4} cos^{-1}[tan( \frac{sin2x}{16} + \frac{x}{8} + C)]$$

 

[Hootenanny]

What's the question? What have you attempted?

~H

 

[JoshHolloway]

The first line is the problem, and each following line is one of the steps I took to solve it. The final line is the solution. Is that right? It sure doesn't look right to me, but I don't know what I am doing wrong.

 

[Hootenanny]

I am a little concerned as to what you have done with respect to this line here;

$$\int \frac{1}{{\color{red} \sqrt{cos(4y)}^{2} + 1}}dy \ = \ \frac{1}{8} \int (cos(2x) + 1)dx$$

Why did you change it into this form? Why did you simply not integrate;

$$\int \frac{1}{cos(4y) + 1}dy$$

In any case

$$\frac{1}{cos(4y) + 1} \neq \frac{1}{ \sqrt{cos(4y)}^{2} + 1}$$

~H

 

[JoshHolloway]

Why is this so:

$$\frac{1}{cos(4y) + 1} \neq \frac{1}{ \sqrt{cos(4y)}^{2} + 1}$$

And I changed it into that form because I thought that would be an easy way to use the arctangent integration technique.

How would I integrate this:
$$\int \frac{1}{cos(4y) + 1}dy$$

 

[JoshHolloway]

Oh, would this work:

$$\int \frac{1}{cos(4y) + 1}dy \ = \ tan^{-1}( \sqrt{cos4y})$$

I reworked the problem that way, and this is what I ended up with:

$$y \ = \ \frac{1}{4} cos^{-1}[tan^{2}( \frac{sin2x}{16} + \frac{x}{8} + C)]$$

Am I getting warmer?

 

[JoshHolloway]

I don't understand why this is so:

$$\frac{1}{cos(4y) + 1} \neq \frac{1}{ \sqrt{cos(4y)}^{2} + 1}$$Isn't this correct?: $$cos(4y) \ = \ \sqrt{cos(4y)}^{2}$$

 

[Hootenanny]

Why is this so:

$$\frac{1}{cos(4y) + 1} \neq \frac{1}{ \sqrt{cos(4y)}^{2} + 1}$$

Sorry I didn't see you had raised it to the power two, my eyes must be getting worse!

How would I integrate this:
$$\int \frac{1}{cos(4y) + 1}dy$$

I think the easiest way to do this is write it in terms of sec².

~H

 

[JoshHolloway]

Does this work:
$$\int \frac{1}{cos(4y) + 1}dy \ = \ tan^{-1}( \sqrt{cos4y})$$

How would you write that in terms of $$sec^{2}$$

 

[JoshHolloway]

Wouldn't you have to use the method of Partial Fractions in order to get it into some form like this in order to write it in terms of sec
$$\frac{A}{cos4y} + 1$$

 

[JoshHolloway]

I have absolutely no idea what you mean by the easiest way would be to get it in terms of sec^2

 

[TD]

I'm having the feeling you're making this harder than necessary.
Remember that (tan(2y))' = 2/(cos²(2y)).

$$\frac{{dy}}{{dx}} = \left( {\cos ^2 x} \right)\left( {\cos ^2 \left( {2y} \right)} \right) \Leftrightarrow \frac{{dy}}{{\cos ^2 \left( {2y} \right)}} = \cos ^2 xdx \Leftrightarrow \frac{1}{2}\tan \left( {2y} \right) = \frac{1}{4}\left( {\sin \left( {2x} \right) + 2x} \right) + C$$

 

[Hootenanny]

Wouldn't you have to use the method of Partial Fractions in order to get it into some form like this in order to write it in terms of sec
$$\frac{A}{cos4y} + 1$$

Yes, using partial fractions;

$$\frac{1}{\cos(4y) +1} = \frac{1}{2}\sec^{2}(2y)$$

Now, you sould find;

$$\int \frac{1}{2}\sec^{2}(2y) \; dy$$

A little easier to solve.

~H

 

[TD]

But we started with sec²(2y), that was the factor cos²(2y) at the rhs...

Perhaps I'm missing something but I think you just went in a circle, from cos²(2y) to the double angle (losing the square) and now back introducing the square, with the half angle!

 

[JoshHolloway]

Wow. Thanks a lot! I really do have a tendency to make things more difficult than they really are. Hey TD, could you check out the last question in the thread I started yesterday. Thanks.

 

[Hootenanny]

But we started with sec²(2y), that was the factor cos²(2y) at the rhs...

Perhaps I'm missing something but I think you just went in a circle, from cos²(2y) to the double angle (losing the square) and now back introducing the square, with the half angle!

Good point, guess my maths is a bit rough, best stick to physics...

~H

 

[TD]

We're all here to help (or to get help), I'd probably make similar mistakes when attempting to answer physics questions

It's nice to see you're helping arround a lot, being a science advisor and homework helper. I remember you from a couple of months back, with a few calculus questions

 

[Hootenanny]

It's nice to see you're helping arround a lot, being a science advisor and homework helper. I remember you from a couple of months back, with a few calculus questions

Hmm, so do I, volumes of revolution and integration by substitution (one where I actually forgot to integrate) once again, my careless mistakes let me down . Thanks again for helping me

~H